a. A uniformy accelerated object is monitored for 10.0sbeginning at a cetain ins

Question

a. A uniformy accelerated object is monitored for 10.0sbeginning at a cetain instant during its motion. If thevelocity of the object at the beginning of the 10.0s perios is 10.0cm/s2, what is its velocity at the end of the 10.0sperios? Assume the acceleration of the object is constant andis equal to 30.0 cm/s2. b. What is the average velcity of the object during the10.0s? c. Use the average velocity to find the distance the objecttravels during the 10.0s period. a. A uniformy accelerated object is monitored for 10.0sbeginning at a cetain instant during its motion. If thevelocity of the object at the beginning of the 10.0s perios is 10.0cm/s2, what is its velocity at the end of the 10.0sperios? Assume the acceleration of the object is constant andis equal to 30.0 cm/s2. b. What is the average velcity of the object during the10.0s? c. Use the average velocity to find the distance the objecttravels during the 10.0s period.

Explanation / Answer

Velocity at the end of 10s is         V = u + at = 10 + 30 *10 = 310 cm/s The diatance travelled in 10 s is       S = ut + 0.5at2          = 10 * 10 + 0.5 *30 * 100          = 100 + 1500          = 1600 cm Average velocity = x /t = 1600cm / 10.0s                                         = 160 cm/s Average velocity = 2* acceleration * distance    distance = V2 - U2/2a                     = (160)2 - (10)2 / 2*30                     = 425cm

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